$\displaystyle (2k+x)^n = 2k^n + ^n\mathrm{C}_1 (2k)^{n-1}x +^n\mathrm{C}_2 (2k)^{n-2} x^2 +^n\mathrm{C}_3 (2k)^{n-3}x^3 +.... $In the binomial expansion of $\displaystyle (2k+x)^n $ wherekis a constant andnis a positive integer, the coefficient of $\displaystyle x^2 $ is equal to the coefficient of$\displaystyle x^3 $.

(a) Prove thatn=6k+2.

Coefficient of $\displaystyle x^2$ = coefficient of $\displaystyle x^3 $

$\displaystyle ^n\mathrm{C}_2(2k)^{n-2} = ^n\mathrm{C}_3(2k)^{n-3} $

$\displaystyle \frac{n!}{ (n-2)!2!} (2k)^{n-2} = \frac{n!}{ (n-3)!3!} (2k)^{n-3} $

I dont know how to simplify this further to get $\displaystyle n=6k+2 $

Any help appreciated,

Thanks.